A consumer researcher wants to test the product claims. The researcher purchases 15 randomly

Question

selected refrigerators and sets them up in a lab. The researcher collects data on how much
electricity each refrigerator uses for 1 month without using the product and then for 1
month with the product. Based on the sample the researcher determines that the average
reduction in energy is approximately normally distributed with a mean of 18%, a standard
deviation of 6%, and a margin of error of 3%.
Based on the data gathered by the consumer researcher, approximately what percent of
refrigerators will reduce their electricity usage by at least 25% when using the product?
[Hint: Use z-score z =
𝑥− 𝜇
𝜎
]
A) P(z) ≈ 1%
B) P(z) ≈ 12%
C) P(z) ≈ 88%
D) P(z) ≈ 99%

Answer 1

To find out the percentage of refrigerators that will reduce their electricity usage by at least 25%, we can use the z-score formula:

\[ z = \frac{x - \mu}{\sigma} \]

where:

• \(x\) is the value we want to find the probability for (25% reduction in this case),
• \(\mu\) is the mean of the distribution (18%),
• \(\sigma\) is the standard deviation (6%).

First, we need to calculate the z-score using the given values:

\[ z = \frac{25% - 18%}{6%} = \frac{7%}{6%} \approx 1.1667 \]

Next, we need to find the probability corresponding to this z-score. We will look this up in a standard normal distribution table or use a calculator that provides cumulative probabilities for the normal distribution.

The cumulative probability for \(z = 1.1667\) is approximately 0.878 (or 87.8%). This value represents the probability that a refrigerator will reduce its electricity usage by less than 25%.

To find the probability that a refrigerator will reduce its electricity usage by at least 25%, we subtract this cumulative probability from 1:

\[ P(Z \geq 1.1667) = 1 - P(Z < 1.1667) \approx 1 - 0.878 = 0.122 \]

This means approximately 12.2% of refrigerators are expected to reduce their electricity usage by at least 25%.

Thus, the closest answer is:

B) P(z) ≈ 12%.